Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define ll long long
#define pii pair<int, int>
#define MAX 100010
using namespace std;

const int INF = 1e9;
void bfs(int s, vector<ll> &d, vector<int> g[]) {
    vector<bool> used(MAX, false);
    d[s] = 0;
    queue<int> que;
    que.push(s);
    while (!que.empty()) {
        int v = que.front();
        que.pop();
        if (used[v]) continue;
        used[v] = true;
        for (int nx : g[v]) {
            if (!used[nx] && d[nx] > d[v] + 1) {
                d[nx] = d[v] + 1;
                que.push(nx);
            }
        }
    }
}

int main() {
    int n, m, s, t;
    cin >> n >> m >> s >> t;
    s--;
    t--;
    vector<int> g[n];
    for (int i = 0; i < m; i++) {
        int x, y;
        cin >> x >> y;
        x--;
        y--;
        g[x].push_back(y);
        g[y].push_back(x);
    }
    vector<ll> ds(MAX, INF), dt(MAX, INF);
    bfs(s, ds, g);
    bfs(t, dt, g);
    vector<int> ca(MAX, 0);
    for (int i = 0; i < n; i++) {
        if (dt[i] == INF) continue;
        ca[dt[i]]++;
    }
    ll ans = 0;
    for (int i = 0; i < n; i++){
        if(dt[i] == INF || ds[i] == INF) continue;
        int x = ds[t] - 2 - ds[i];
        if (x >= 0) ans += ca[x];
    }
    cout << ans << endl;
    return 0;
}